MathDB

9

Part of 2018 Irish Math Olympiad

Problems(1)

a_{n+1} = a^2_{n} + 2018 for n>=1, exists at most one perfect cube

Source: Irmo 2018 p2 q9

9/16/2018
The sequence of positive integers a1,a2,a3,...a_1, a_2, a_3, ... satisfies an+1=an2+2018a_{n+1} = a^2_{n} + 2018 for n≥1n \ge 1. Prove that there exists at most one nn for which ana_n is the cube of an integer.
perfect cubeSequencerecurrence relationnumber theory